Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta thấy: \(f\left(x\right)=\frac{x^3}{1-3x+x^2}\)
\(f\left(1-x\right)=\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+\left(1-x\right)^2}\)\(=\frac{\left(1-x\right)^3}{x^2-3x+1}\)
\(f\left(x\right)+f\left(1-x\right)=\frac{x^3+\left(1-x\right)^3}{x^2-3x+1}\)=1
Do đó: \(f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)=1\)
\(f\left(\frac{2}{2012}\right)+f\left(\frac{2010}{2012}\right)=1\)
....
\(f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)=1\)
=>A=1+1+1+...+1+\(f\left(\frac{1006}{2012}\right)\)=\(\frac{2009}{2}\)
(1005 số 1)
bn ơi cho mình hỏi dòng thứ 2 á tại sao \(\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+\left(1-x\right)^2}=\frac{\left(1-x\right)^3}{x^2-3x+1}\)
Đễ dàng chưng minh được
\(f\left(1-x\right)=1-f\left(x\right)\)
\(\Rightarrow f\left(1-x\right)+f\left(x\right)=1\)
\(\Rightarrow A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+\left[f\left(\frac{2}{2012}\right)+f\left(\frac{2010}{2012}\right)\right]+...+\left[f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)\right]+f\left(\frac{1006}{2012}\right)\)
\(=1005+f\left(\frac{1006}{2012}\right)\)
Làm nôt
\(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
Ta có \(f\left(x\right)+f\left(1-x\right)=1\) khi đó
\(A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+...+\left[f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)\right]+f\left(\frac{1006}{2012}\right)\)
\(=1+1+...+1+f\left(\frac{1}{2}\right)=1005+\frac{\left(\frac{1}{2}\right)^3}{1-3.\frac{1}{2}+3.\left(\frac{1}{2}\right)^2}=1005+\frac{1}{2}=\frac{2011}{2}\)
Ta có: \(F\left(x\right)=\frac{x^3}{1-3x+3x^2}\)
\(\Leftrightarrow F\left(1-x\right)=1-\frac{x^3}{1-3x+3x^2}\)
\(=\frac{1-3x+3x^2-x^3}{1-3x+3x^2}\)
\(=\frac{\left(1-x\right)^3}{1-3x+3x^2}\)
Ta có: \(F\left(x\right)+F\left(1-x\right)\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3x+3x^2}\)
\(=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
\(\Leftrightarrow F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)=1\)
...
\(F\left(\frac{1005}{2012}\right)+F\left(\frac{1007}{2012}\right)=1\)
Do đó: \(A=F\left(\frac{1}{2012}\right)+F\left(\frac{2}{2012}\right)+...+F\left(\frac{2010}{2012}\right)+F\left(\frac{2011}{2012}\right)\)
\(=\left[F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)\right]+\left[F\left(\frac{2}{2012}\right)+F\left(\frac{2010}{2012}\right)\right]+...+F\left(\frac{1006}{2012}\right)\)
\(=1+1+...+F\left(\frac{1}{2}\right)\)
\(=1005+\left[\left(\frac{1}{2}\right)^3:\left(1-3\cdot\frac{1}{2}+3\cdot\frac{1}{4}\right)\right]\)
\(=1005+\left[\frac{1}{8}:\left(1-\frac{3}{2}+\frac{3}{4}\right)\right]\)
\(=1005+\left(\frac{1}{8}:\frac{1}{4}\right)\)
\(=1005+\frac{1}{2}=\frac{2011}{2}\)
......................?
mik ko biết
mong bn thông cảm
nha ................
Ta có : \(\left(x-1\right)f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}\) (1)
Thay x bởi \(\frac{1}{x}\)thì đẳng thức thành :
\(\left(\frac{1}{x}-1\right)f\left(\frac{1}{x}\right)+f\left(x\right)=\frac{1}{\frac{1}{x}-1}\)
Hay : \(\frac{1-x}{x}f\left(\frac{1}{x}\right)+f\left(x\right)=\frac{x}{1-x}\) (2)
Nhân \(\frac{1-x}{x}\)vào hai vế của (1), ta được :
\(\frac{-x^2+2x-1}{x}f\left(x\right)+\frac{1-x}{x}f\left(\frac{1}{x}\right)=-\frac{1}{x}\) (3)
Lấy (2) trừ đì (3) theo từng vế, ta được :
\(\left[1-\frac{-x^2+2x-1}{x}\right]f\left(x\right)=\frac{x}{1-x}+\frac{1}{x}\)
Suy ra : \(f\left(x\right)=\frac{1}{1-x}\)
Xét \(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3+3x+3-6x+3x^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}\)
\(=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
Thay vào ta tính được:
\(A=\left[f\left(\frac{1}{2020}\right)+f\left(\frac{2019}{2020}\right)\right]+...+\left[f\left(\frac{1009}{2020}\right)+f\left(\frac{1011}{2020}\right)\right]+f\left(\frac{1010}{2020}\right)\)
\(A=1+...+1+f\left(\frac{1010}{2020}\right)\) (với 1009 số 1)
\(A=1009+f\left(\frac{1}{2}\right)=1009+\frac{\left(\frac{1}{2}\right)^3}{1-3\cdot\frac{1}{2}+3\cdot\left(\frac{1}{2}\right)^2}\)
\(A=1009+\frac{1}{2}=\frac{2019}{2}\)
Vậy \(A=\frac{2019}{2}\)